Oxidation is the loss of electrons. In Mn(s) the manganese is in the 0 oxidation state, while Mn2+(aq) is +2. To go from 0 to +2, manganese must lose two electrons. That loss of electrons is shown on the product side of the half-reaction, giving Mn(s) → Mn2+ + 2e−. That’s why this is the oxidation half-reaction for manganese in this process. The reverse reaction Mn2+ + 2e− → Mn(s) would be a reduction. The option with Ti involves a different element, not manganese. And “None” isn’t correct because there is a valid oxidation half-reaction here.

The key idea is that the overall cell potential comes from the difference between the standard reduction potentials of the two half-reactions: E°cell = E°cathode − E°anode. To get the largest positive value, place the half-reaction with the most positive E° as the cathode and the one with the most negative E° as the anode. The strongest positive standard potential is for fluorine reducing to fluoride, and the strongest negative standard potential is for lithium metal forming Li+. When you take that large positive potential and subtract the large negative potential, you get a very large positive E°cell—on the order of several volts, near the upper limit of what standard potentials can give. That explains why the largest approximate value is around six volts, far larger than the other options.

In a concentration cell the driving force is the difference in concentrations of the same redox couple on the two sides. The Nernst equation shows that each half-cell’s potential depends on the ratio of activities (essentially concentrations) of its oxidized and reduced forms. When you have identical couples on both sides, the overall cell potential reduces to a term that depends only on the ratio of those activities between sides. The sign comes from which side has the higher activity of the oxidized species: the electrode with the larger oxidized-form activity tends to oxidize (becomes the anode), and the other side tends to reduce (becomes the cathode), giving a positive Ecell in the defined direction if the cathode side has the higher oxidized-form activity. The ratio of activities on the two electrodes is therefore what sets the sign. Other factors like solvent viscosity, external pressure, or electrode surface area do not determine the EMF sign in this ideal description; they can influence kinetics or practical performance but not the thermodynamic driving force.

This is about how the mass of material deposited during electroplating or a redox process relates to the total electric charge passed, using Faraday’s law. The amount of substance deposited is directly proportional to the charge, with the proportionality constant built from the molar mass, the number of electrons needed to deposit one atom (or ion), and Faraday’s constant. Specifically, for a deposit with molar mass M that requires n electrons per atom, the charge needed to deposit one mole of atoms is nF. Therefore, the number of moles deposited for a total charge Q is Q divided by nF. Multiplying by the molar mass M converts that to mass, giving m = M Q /(n F). This shows why the linear form with Q in the numerator and n and F in the denominator correctly describes the relationship. If you tried to put n in the numerator or square F or use Q^2, you’d violate the linear dependence on charge and the units, leading to an incorrect amount. The correct expression is m = (M Q)/(n F).